Question

I’m trying to render the following HTML code using xslt:

<foo:bar href="[MYURL]"/>

I created the following template:

<xsl:template name="displaySomeLink">
    <xsl:param name="url"/>
    <foo:bar href="{$url}"></foo:bar>
</xsl:template>

Which I’m planning to call like:

<xsl:call-template name="displaySomeLink">
    <xsl:with-param name="url" select="[MYURLSELECTIONLOGICS]"/>
</xsl:call-template>

However, my xslt parser indicates justly that the foo prefix is not defined, and as the foo prefix doesn’t signify anything at xslt level, I don’t want to register it. Is there any way I can escape a character in a tagname?

(Of course I can easily do some other workaround, like concatenating a string which will output the same.)

Answer 1

Just define the foo namespace prefix in the XSLT stylesheet:

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
                xmlns:foo="http://foo.foo/foo"
                ...>

or

<xsl:template name="whatever" xmlns:foo="http://foo.foo/foo">
  ...

or

<xsl:template name="whatever">
   ...
   <foo:bar xmlns:foo="http://foo.foo/foo" />
   ...

You are correct in that the foo namespace prefix and its namespace have no meaning in XSLT, but neither have the default namespace “” or any other namespace except for the http://www.w3.org/1999/XSL/Transform namespace (usually represented by the xsl prefix).

The XSLT processor will just spew out elements with whatever namespace they are defined in.